Let R_n be the quotient ring and let S_n be the associated algebra introduced in the paper. Let O_i, for i∈0,…,k, be the orbits in R_n under the “multiply-by-X” F_2[X]-action. Orbit-degree decomposition conjecture. Then S_n splits as a direct sum of fields of degree | O_i| for all i∈0,…,k. The preceding proposition gives a related decomposition into isomorphic algebras and fields whose degrees divide the orbit lengths. The stated equality of every field degree with the corresponding orbit size is presented as an open conjecture, with no resolution supplied in the source.

The orbit-degree decomposition conjecture for finite-field algebras

About 27 years old · traced to 1999

Let $R_n$ be the quotient ring and let $S_n$ be the associated algebra introduced in the paper. Let $\mathcal O_i$ , for $i\in\{0,\dots,k\}$ , be the orbits in $R_n$ under the “multiply-by- $X$ ” $\mathbb F_2[X]$ -action.

Orbit-degree decomposition conjecture. Then $S_n$ splits as a direct sum of fields of degree $|\mathcal O_i|$ for all $i\in\{0,\dots,k\}$ .

The preceding proposition gives a related decomposition into isomorphic algebras and fields whose degrees divide the orbit lengths. The stated equality of every field degree with the corresponding orbit size is presented as an open conjecture, with no resolution supplied in the source.

References

Primary source

Laurent Bartholdi, “Lamps, Factorizations and Finite Fields”, arXiv:math/9910056 (1999).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Open

A posted argument claims a complete proof for every allowed size, but no independent verification is recorded.

The conjecture asserts that each field component of $S_n$ has degree equal to the size of the corresponding orbit in $R_n$ . The source presents this equality as open, beyond a weaker decomposition into related algebras and fields.

Posted attempt

An argument claims the result for the broader family $S_f=\mathbb F_q[X]/(L_f)$ , by identifying Frobenius roots with $R_f$ as modules, matching orbit lengths with Frobenius-orbit lengths, and applying separability and the Chinese remainder theorem. Specializing to $q=2$ and $f(T)=T^n+T+1$ is claimed to prove the conjecture for every $n\ge2$ , including the zero orbit; the argument has not been independently verified.

Current status (as of August 2026): A complete proof has been claimed for the conjecture, but it remains unverified; absent confirmation, the conjecture is not settled.

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Theorem. Let $q$ be any prime power and let

f(T)=a_0+a_1T+\cdots+a_dT^d\in\mathbf F_q[T], \qquad d\ge1,\quad a_0a_d\ne0.

Put

R_f=\mathbf F_q[T]/(f),\qquad L_f(X)=\sum_{i=0}^d a_iX^{q^i},\qquad S_f=\mathbf F_q[X]/(L_f).

If $\mathcal O$ ranges over the orbits of multiplication by $T$ on the underlying set of $R_f$ , then

S_f\cong\prod_{\mathcal O}\mathbf F_{q^{|\mathcal O|}}.

Proof. Because $a_0\ne0$ , the class of $T$ is a unit in the finite ring $R_f$ . Let $m$ be its multiplicative order. Then

T^m-1=f(T)h(T)

for some $h\in\mathbf F_q[T]$ . Write $F(u)=u^q$ for Frobenius on $K=\mathbf F_{q^m}$ . The roots of $L_f$ are exactly $V=\ker f(F)$ . Every such root belongs to $K$ , since

(F^m-I)u=h(F)f(F)u=0.

The normal-basis theorem gives a basis

\theta,F\theta,\ldots,F^{m-1}\theta

of $K$ over $\mathbf F_q$ . Consequently,

\mathbf F_q[T]/(T^m-1)\longrightarrow K, \qquad[g]\longmapsto g(F)\theta,

is an isomorphism of $\mathbf F_q[T]$ -modules. Therefore

V\cong\ker\left(f:\mathbf F_q[T]/(fh)\to\mathbf F_q[T]/(fh)\right) =h\mathbf F_q[T]/(fh).

Multiplication by $h$ induces

\mathbf F_q[T]/(f)\xrightarrow{\sim}h\mathbf F_q[T]/(fh), \qquad[g]\longmapsto[gh].

Indeed, $fh\mid gh$ implies $f\mid g$ , so this map is injective and plainly surjective. No coprimality between $f$ and $h$ is needed.

Thus $V\cong R_f$ as $\mathbf F_q[T]$ -modules, with $T$ acting as Frobenius on $V$ and as multiplication by $T$ on $R_f$ . Their orbit multisets therefore coincide. Since

L_f'(X)=a_0\ne0,

the polynomial $L_f$ is separable. Its irreducible factors correspond exactly to Frobenius orbits of roots, and their degrees equal the corresponding orbit lengths. The Chinese remainder theorem now gives

S_f\cong\prod_{\mathcal O}\mathbf F_{q^{|\mathcal O|}}.

Taking $q=2$ and $f(T)=T^n+T+1$ yields

L_f(X)=X^{2^n}+X^2+X

and proves the conjectured decomposition

\boxed{\displaystyle S_n\cong\prod_{\mathcal O}\mathbf F_{2^{|\mathcal O|}}}

for every $n\ge2$ , including the zero orbit. This orbit-degree result makes no assertion about multiplicative primitivity of individual roots.